You do not need currentDistance here.

Once we find the minDistance to be equal to 1, we can just return because it can never go down below that. This will help in saving some time as well if the first two words itself are word1 and word2.

Don't need either currentDistance or to keep track of both indices:

min_dist = len(words)
curr_word, idx = None, 0
for i, w in enumerate(words):
    if w not in (word1, word2): continue
    if curr_word and w != curr_word:
        min_dist = min(min_dist, i - idx)
    curr_word, idx = w, i
return min_dist

Check words[i] is word1 or word2, else continue. This can save a little.

Why the 2nd approach has Time complexity: O(N⋅M) and not O(N)? Why "the total length of two input words." influences the time complexity here?

My python submission

out = first = second = float('inf')
        for i,word in enumerate(words):
            if word == word1: first = i
            elif word == word2: second = i
            out = min(abs(first - second), out)
        return out

This easy question in interview will turn into a super hard problem by maintaining list of indices for each word and performing a binary search and so on :-)

Just for fun, here is a more robust solution if we consider a few more edge cases (word 1 or word 2 are not in the list, etc).

    public int shortestDistance(String[] words, String word1, String word2) {
        if ((words == null) || (words.length == 0)) 
            return -1;
        
        int index_w1=-1, index_w2=-1;
        int shortestDistance = Integer.MAX_VALUE;
        for (int i=0; i<words.length; i++) {
            if (words[i].equals(word1))
                index_w1 = i;            
            if (words[i].equals(word2))
                index_w2 = i;
            
            // We found a new path between w1 and w2
            if ((index_w1!=-1) && (index_w2!=-1)) {
                shortestDistance = Math.min(shortestDistance, Math.abs(index_w1-index_w2));
            }            
        }
        
        return shortestDistance == Integer.MAX_VALUE? -1 : shortestDistance;
    }

def shortestDistance(self, words, word1, word2):
    """
    :type words: List[str]
    :type word1: str
    :type word2: str
    :rtype: int
    """
    shortest = len(words)+1
    index = None
    found = None
    for i, word in enumerate(words):
        if word == word1 or word == word2:
            if found is not None:
                if word != found:
                    shortest = min(shortest, i - index)
            found = word
            index = i
    return shortest

There an error for the one pass solution minDistance = Math.min(minDistance, Math.abs(i1 - i1)); have to be minDistance = Math.min(minDistance, Math.abs(i1 - i2));